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Proof step 89 from proof bitsand-cast (previous step)

∃ bits3, bits2. bits0 = bits3 + bits2 ∧ # bits2 = 2 + # bits1, sgn(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1) = boolvar ' + boolvar0 ' + bits1, n = # fillsgn(n, mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1), 2 + # bits1 < # bits0, n0 + 1 + # bits1 < n ⊦ lastn(n, fillsgn(n, bitsand(boolvar ' + boolvar0 ' + bits1, bits0))) = lastn(n, fillsgn(n, bitsand(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1, bits0))), n0 = 0, boolvar ↔ boolvar0, bits0 = [], boolvar ∧ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 90:
# bits = 2 + # bits1, sgn(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1) = boolvar ' + boolvar0 ' + bits1, n = # fillsgn(n, mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1), n0 + 1 + # bits1 < n ⊦ bits2 = [], lastn(n, fillsgn(n, bitsand(boolvar ' + boolvar0 ' + bits1, bits2 + bits))) = lastn(n, fillsgn(n, bitsand(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1, bits2 + bits))), n0 = 0, boolvar ↔ boolvar0, bits2 + bits = [], boolvar ∧ boolvar0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;